OT - But I am buiding a tool

Can anyone help me with a vector force problem or point me to a forum that might give me a formula. I am building a dump bed trailer and I need to know the force required for the hydraulic cylinder to start to lift the load.

The trailer bed is 4� x 8� and will have a maximum load of 3000 lbs evenly loaded. The pivot point of the bed is at the top of triangle �C� in the drawing and is 7 feet from the front and 1 foot from the back of the trailer. A hydraulic cylinder 22.25� closed length will anchor at points �A� and �B�. The distance from point �B� to pivot point �C� is 37.29�. The distance from point �A� to pivot point �C� is 16.23�. Angle �ABC� IS 12.18 degrees. Angle �ACB� is 16.81 degrees. Angle �BAC� is 151.01 degrees. I know I could just oversize the cylinder and it would work but I would really like to have a formula to calculate it so I can play with different cylinder diameters and lengths as well as different mounting points on the trailer.

Trust me. :)

--->Paul

Two easy and effective ways to reduce the loads are to either lower point B, or move point A to the left.

Right now, your angle is so tight you'd cut the forces in half by just moving point B down 4". Try the math and see.

Since my Physics class is asking help while I type this, I will need to attack this later.

The subject that you want is Statics. You might try googling for a website. Basically you sum the forces and the moments at each point to find the loading.

HTH

Cliff (VA)

First we load balance. Since it's evenly distributed along the bed, we can model it as centered in the bed.

Now we measure the torque imparted around the hing, point C. This gives a torque around the hinge of 3000lbs x 3 ft = 9000 lb/ft.

Next we counter that by point A. We just work with the straight up load to counter the bed torque around the hinge. Distance is 16.23 inches, so I'll round that to 16", which is 1.3 feet. 9000 lb/ft / 1.3 = 6923 lbs straight up at point A.

Don't flip out just yet. Yes, 6923 is more than the 3000 lb load. Remember, this is an offset load. With the load sitting on just A and C, it's trying to flip over to the left. The ram is going to be pushing up the load, and countering the leverage, effectively trying to rip the hinge out of the truck. In fact, the hinge is going to see 3923 lbs of force trying to rip it upwards.

Now some trig to figure out the force on the cylinder.

The angle I'll use you give as 12.18. Sine is opposite over hypotenuse. In this case, the hyponenuse is the cylinder. Manipulate and get hyponenuse = opposite / sine angle.

Opposite is 6923 lbs. Sine of 12.18 = 0.19
Hypotenuse = 36,436 lbs (6923 / 0.19)

That's the force the cylinder's got to exert to get that load just lifting, 36, 436 lbs force.

So lets do it that way and see what happens. We'll treat the load on the left of the hinge seperately from the load on the right.

We take the 3000 lb spread out uniformly and figure out the load per foot on the bed. 3000 lb / 8 ft = 375 lb/ft along the bed.

On the right, we've got 1 foot of bed with 375 lbs. Center that load and it's six inches to the right of hinge C, or 0.5 foot. Figure the torque. 375 lb * 0.5 ft = 187.5 lb/ft of torque, clockwise.

On the left side, we've got 375 lb/ft * 7 ft = 2625 lb. And being a 7 foot bed on the left, that load would be centered 3.5 feet to the left of the hinge pin C. Figure that torque. 2625 lb * 3.5 ft = 9187.5 lb/ft of torque, counter-clockwise.

Now add the two torques together.

9187.5 lb/ft counter-clockwise + 187.5 lb/ft clockwise = 9000 lb/ft counter-clockwise.

Remember, the addition I just did is of a counter force, so it looked like subtraction as the clockwise force is countering the counter-clockwise force, and only the difference between the two forces remains.

The 9000lb figure would actually be 15000lbs at point C. At the left end of the dump bed you would need 3000lbs of lift to move a 3000lbs load at zero resistance. So if you move in 5ft at point C then your lifting 5 x 3000 = 15000lbs at point C and not 9000lbs ?

Terry since you will being buying pre-made cylinder(s) in common Known open lengths, I would start with the dumpbed open at maximum dump degrees then run 5� off perpendicular to the frame would locate the cylinder to frame attachment (point C) then calculate the cylinder diameter size needed to over come lift force at that point.

Besure and do not forget to add the friction resistance it takes to get the load moving.

T_Bone

No, Nolan has it right. You have to calculate the moment generated by the load, which you do by figuring out the center of the the load and using the distance from that point to your pivot as the momemt arm. In this case, the momemt arm is 3 ft, resulting in a momement of 9000 ft-lbs.

If this seems counterintuitive, just consider the case where the pivot is directly under the center of the load. In this case, the moment arm is zero, resulting in a zero moment. Meaning that the load is balanced on the pivot, and only a small force is required to rotate the load around the pivot. A teeter-totter, if you will.

Nolan and I came up with different answers due to round-off error and because we used different values for the sine of 21 degrees. But we used exactly the same method to solve the problem: calculate moment, calculate the vertical force component, calculate the total force.

First we determine the moment generated by the load. The 3000 lb load is centered at a point 3 ft from the pivot, which gives us a moment of 9000 ft-lbs.

M = Fd = 3000 x 3 = 9000

Now the vertical component of the force generated by the cylinder has a moment arm of 16.23 inches (1.35 ft), so we need a force of 6650 lbs to generate the 9000 ft-lb moment.

F = M/d = 9000/1.35 = 6670

The vertical component of the force generated by the cylinder is going to be the total force of the cylinder times the sine of the angle ABC. (sine 12.2 degrees = .211). So the cylinder needs to generate a force of about 31,600 lbs.

F = Fv / sin(12.2) = 6670/.211 = 31,600

But this is just to support the weight of the load. You will need more force to actually lift the load, both to overcome friction and to accelerate the load upward. So maybe 40,000 lbs.

The determining factors on the size of cylinder you need are the distance from the pivot point and angle ABC. Increasing either of these will reduce the force required to lift the load. Of course, as you've probably already figured out, it is difficult to increase one without decreasing the other.

To have a hydraulic cylinder represented by line AB as drawn, and have it actually lift the bed, pivot point C would have to be on the LEFT side of the diagram. To do otherwise, when cylinder AB extends, you lose the advantage of leverage.

Think of your trailer as a lever, with the fulcrum at point C. To attempt to lift a load with the average weight at the center of the trailer, by attaching your cylinder to point A as illustrated is, in effect, shortening the length of your lever...as point A is where your upward force vector would be. But if you reversed the direction of line AB, to where point A was on the frame and point B was on the bed, your upward force vector would now be at point B...and your "lever" length from fulcrum C to new lift point B is longer...thus requiring less force to lift the load.

I hope I expressed myself clearly here. If you examine most hydraulic dump beds, such as grain beds, the cylinders are most often anchored to the truck frame at point A, in relation to the fulcrum/pivot point C...and anchored to the bed at point B...so that when the cylinder extends, the left side in your diagram would be the one being raised.

As I said, I dropped out of engineering school after a year, so I can't help you with the amount of force needed to dump the trailer. But I do know the laws of physics enough to understand that lifting at point B, rather than point A, takes less effort because you've lengthened your [effective] lever.

You eat an elephant one bite at a time.

• Hobby farm

I will agree the cylinder is located in about the worst possible location. Need a longer one closer to the end - or flip it around so it is pushing to the front. Or you get into some _serious_ forces.

--->Paul